What is the extraneous solution to these equations? $\dfrac{x^2}{x + 10} = \dfrac{x + 2}{x + 10}$
Explanation: Multiply both sides by $x + 10$ $ \dfrac{x^2}{x + 10} (x + 10) = \dfrac{x + 2}{x + 10} (x + 10)$ $ x^2 = x + 2$ Subtract $x + 2$ from both sides: $ x^2 - (x + 2) = x + 2 - (x + 2)$ $ x^2 - x - 2 = 0$ Factor the expression: $ (x + 1)(x - 2) = 0$ Therefore $x = -1$ or $x = 2$ The original expression is defined at $x = -1$ and $x = 2$, so there are no extraneous solutions.